Answer
$β=0.8186$
Power of the test: 0.1814
Work Step by Step
$z_α=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33.
$p ̂=p_0+z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.65+2.33\sqrt {\frac{0.65(1-0.65)}{100}}=0.761$
$β=P(Type~II~error)=P(p ̂\lt0.761~given~that~p=0.72)$
$β=P(z\lt\frac{0.761-0.72}{\sqrt {\frac{0.72(1-0.72)}{100}}})=P(z\lt0.91)=0.8186$
Power of the test: $1-β=0.1814$
